00:01
Hello, in the question it is given, a simple used shaped container that is consider one u shaped container here.
00:18
And the mercury, this u shaped container contains mercury to the same level in both of its arms.
00:27
So consider one mercury that is consider mercury for same height, it be both the arms.
00:46
This is the mercury which is poured into this new shape vessel.
00:55
Now what happens is suppose water is poured to a height of 13 .6 centimeter in one arm.
01:03
So i am considering this arm here and water is poured up to one height of 13 .6 centimeter.
01:13
So in blue color it is water and in green color it is mercury.
01:20
So height of the water column is let us consider it as that is h and height of water poured h is equal to 13 .6 cm.
01:36
So due to this water insertation and there will be some pressure exerted on the mercury as well and there will be some rays in mercury level in the another arm.
01:49
So we need to find out how much will be the rise in the rise in one.
01:52
Mercury level in other arm that is h dash value we we need to find out and density of mercury is given that is 13 .6 into 10 to the power 3 kg per meter cube and density of water is thousand kg per meter cube or we can write 10 to the power 3 kg per meter cube so since the length of the mercury column in two arms are equal height to which water is poured in one arm is 13 .6 cm.
02:28
So, h -dash that we need to calculate, that is rise in mercury in another column.
02:35
Now, so pressure due to water at one arm, that is, let us consider it as p1, that is equal to pressure on hg on another arm, that is, p1 must be equal to p2...